Term Rewriting System R:
[x, y]
p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

LE(s(x), s(y)) -> LE(x, y)
MINUS(x, y) -> IF(le(x, y), x, y)
MINUS(x, y) -> LE(x, y)
IF(false, x, y) -> MINUS(p(x), y)
IF(false, x, y) -> P(x)

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

LE(s(x), s(y)) -> LE(x, y)

Rules:

p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

LE(s(x), s(y)) -> LE(x, y)

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
LE(x1, x2) -> LE(x1, x2)
s(x1) -> s(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`           →DP Problem 3`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

Rules:

p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳Narrowing Transformation`

Dependency Pairs:

IF(false, x, y) -> MINUS(p(x), y)
MINUS(x, y) -> IF(le(x, y), x, y)

Rules:

p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

MINUS(x, y) -> IF(le(x, y), x, y)
three new Dependency Pairs are created:

MINUS(0, y'') -> IF(true, 0, y'')
MINUS(s(x''), 0) -> IF(false, s(x''), 0)
MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Narrowing Transformation`

Dependency Pairs:

MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))
MINUS(s(x''), 0) -> IF(false, s(x''), 0)
IF(false, x, y) -> MINUS(p(x), y)

Rules:

p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

IF(false, x, y) -> MINUS(p(x), y)
two new Dependency Pairs are created:

IF(false, 0, y) -> MINUS(0, y)
IF(false, s(x''), y) -> MINUS(x'', y)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 5`
`                 ↳Instantiation Transformation`

Dependency Pairs:

MINUS(s(x''), 0) -> IF(false, s(x''), 0)
IF(false, s(x''), y) -> MINUS(x'', y)
MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))

Rules:

p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF(false, s(x''), y) -> MINUS(x'', y)
two new Dependency Pairs are created:

IF(false, s(x''''), 0) -> MINUS(x'''', 0)
IF(false, s(x''''), s(y'''')) -> MINUS(x'''', s(y''''))

The transformation is resulting in two new DP problems:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 6`
`                 ↳Argument Filtering and Ordering`

Dependency Pairs:

IF(false, s(x''''), 0) -> MINUS(x'''', 0)
MINUS(s(x''), 0) -> IF(false, s(x''), 0)

Rules:

p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF(false, s(x''''), 0) -> MINUS(x'''', 0)

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
IF(x1, x2, x3) -> x2
s(x1) -> s(x1)
MINUS(x1, x2) -> x1

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 8`
`                 ↳Dependency Graph`

Dependency Pair:

MINUS(s(x''), 0) -> IF(false, s(x''), 0)

Rules:

p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 7`
`                 ↳Argument Filtering and Ordering`

Dependency Pairs:

IF(false, s(x''''), s(y'''')) -> MINUS(x'''', s(y''''))
MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))

Rules:

p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF(false, s(x''''), s(y'''')) -> MINUS(x'''', s(y''''))

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
IF(x1, x2, x3) -> x2
s(x1) -> s(x1)
MINUS(x1, x2) -> x1

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 9`
`                 ↳Dependency Graph`

Dependency Pair:

MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))

Rules:

p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes